1046
11.10
Applications of Trigonometry
Parametric Equations
As we have seen in Exercises 53 - 56 in Section 1.2, Chapter 7 and most recently in Section 11.5,
there are scores of interesting curves which, when plotted in the xy-plane, neither represent y as a
function of x nor x as a function of y. In this section, we present a new concept which allows us
to use functions to study these kinds of curves. To motivate the idea, we imagine a bug crawling
across a table top starting at the point O and tracing out a curve C in the plane, as shown below.
y
P (x, y) = (f (t), g(t))
5
4
Q
3
2
1
O
1
2
3
4
5
x
The curve C does not represent y as a function of x because it fails the Vertical Line Test and it
does not represent x as a function of y because it fails the Horizontal Line Test. However, since
the bug can be in only one place P (x, y) at any given time t, we can define the x-coordinate of P
as a function of t and the y-coordinate of P as a (usually, but not necessarily) different function of
t. (Traditionally, f (t) is used for x and g(t) is used for y.) The independent variable t in this case
is called a parameter and the system of equations
x = f (t)
y = g(t)
is called a system of parametric equations or a parametrization of the curve C.1 The
parametrization of C endows it with an orientation and the arrows on C indicate motion in the
direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to
the left, then loops back around to cross its path2 at the point Q and finally heads off into the
first quadrant. It is important to note that the curve itself is a set of points and as such is devoid
of any orientation. The parametrization determines the orientation and as we shall see, different
parametrizations can determine different orientations. If all of this seems hauntingly familiar,
it should. By definition, the system of equations {x = cos(t), y = sin(t) parametrizes the Unit
Circle, giving it a counter-clockwise orientation. More generally, the equations of circular motion
{x = r cos(ωt), y = r sin(ωt) developed on page 732 in Section 10.2.1 are parametric equations
which trace out a circle of radius r centered at the origin. If ω > 0, the orientation is counterclockwise; if ω < 0, the orientation is clockwise. The angular frequency ω determines ‘how fast’ the
1
Note the use of the indefinite article ‘a’. As we shall see, there are infinitely many different parametric representations for any given curve.
2
Here, the bug reaches the point Q at two different times. While this does not contradict our claim that f (t) and
g(t) are functions of t, it shows that neither f nor g can be one-to-one. (Think about this before reading on.)
11.10 Parametric Equations
1047
π
π
object moves around the circle. In particular, the equations x = 2960 cos 12
t , y = 2960 sin 12
t
that model the motion of Lakeland Community College as the earth rotates (see Example 10.2.7 in
Section 10.2) parameterize a circle of radius 2960 with a counter-clockwise rotation which completes
one revolution as t runs through the interval [0, 24). It is time for another example.
x = t2 − 3
Example 11.10.1. Sketch the curve described by
for t ≥ −2.
y = 2t − 1
Solution. We follow the same procedure here as we have time and time again when asked to graph
anything new – choose friendly values of t, plot the corresponding points and connect the results
in a pleasing fashion. Since we are told t ≥ −2, we start there and as we plot successive points, we
draw an arrow to indicate the direction of the path for increasing values of t.
y
t
−2
−1
0
1
2
3
x(t) y(t) (x(t), y(t))
1 −5
(1, −5)
−2 −3
(−2, −3)
−3 −1
(−3, −1)
−2
1
(−2, 1)
1
3
(1, 3)
6
5
(6, 5)
5
4
3
2
1
−2−1
−1
1
2
3
4
5
6
x
−2
−3
−5
The curve sketched out in Example 11.10.1 certainly looks like a parabola, and the presence of
2
2
the
t 2term in the equation x = t − 3 reinforces this hunch. Since the parametric equations
x = t − 3, y = 2t − 1 given to describe this curve are a system of equations, we can use the
technique of substitution as described in Section 8.7 to eliminate the parameter t and get an
equation involving just x and y. To do so, we choose to solve the equation y = 2t − 1 for t to
2
2 − 3 yields x = y+1
get t = y+1
.
Substituting
this
into
the
equation
x
=
t
− 3 or, after some
2
2
rearrangement, (y + 1)2 = 4(x + 3). Thinking back to Section 7.3, we see that the graph of this
equation is a parabola with vertex (−3, −1) which opens to the right, as required. Technically
speaking, the equation (y + 1)2 = 4(x + 3) describes the entire parabola, while the parametric
equations x = t2 − 3, y = 2t − 1 for t ≥ −2 describe only a portion of the parabola. In this case,3
we can remedy this situation by restricting the bounds on y. Since the portion of the parabola we
want is exactly the part where y ≥ −5, the equation (y + 1)2 = 4(x + 3) coupled with the restriction
y ≥ −5 describes the same curve as the given parametric equations. The one piece of information
we can never recover after eliminating the parameter is the orientation of the curve.
Eliminating the parameter and obtaining an equation in terms of x and y, whenever possible,
can be a great help in graphing curves determined by parametric equations. If the system of
parametric equations contains algebraic functions, as was the case in Example 11.10.1, then the
usual techniques of substitution and elimination as learned in Section 8.7 can be applied to the
3
We will have an example shortly where no matter how we restrict x and y, we can never accurately describe the
curve once we’ve eliminated the parameter.
1048
Applications of Trigonometry
system {x = f (t), y = g(t) to eliminate the parameter. If, on the other hand, the parametrization
involves the trigonometric functions, the strategy changes slightly. In this case, it is often best
to solve for the trigonometric functions and relate them using an identity. We demonstrate these
techniques in the following example.
Example
x
1.
y
x
2.
y
11.10.2. Sketch the curves described by the following
= t3
x =
for
−1
≤
t
≤
1
3.
= 2t2
y =
= e−t
x =
for t ≥ 0
4.
= e−2t
y =
parametric equations.
sin(t)
for 0 < t < π
csc(t)
1 + 3 cos(t)
for 0 ≤ t ≤
2 sin(t)
3π
2
Solution.
1. To get a feel for the curve described by the system x = t3 , y = 2t2 we first sketch the
graphs of x = t3 and y = 2t2 over the interval [−1, 1]. We note that as t takes on values
in the interval [−1, 1], x = t3 ranges between −1 and 1, and y = 2t2 ranges between 0
and 2. This means that all of the action is happening on a portion of the plane, namely
{(x, y) | − 1 ≤ x ≤ 1, 0 ≤ y ≤ 2}. Next, we plot a few points to get a sense of the position
and orientation of the curve. Certainly, t = −1 and t = 1 are good values to pick since
these are the extreme values of t. We also choose t = 0, since that corresponds to a relative
minimum4 on the graph of y = 2t2 . Plugging in t = −1 gives the point (−1, 2), t = 0 gives
(0, 0) and t = 1 gives (1, 2). More generally, we see that x = t3 is increasing over the entire
interval [−1, 1] whereas y = 2t2 is decreasing over the interval [−1, 0] and then increasing
over [0, 1]. Geometrically, this means that in order to trace out the path described by the
parametric equations, we start at (−1, 2) (where t = −1), then move to the right (since x is
increasing) and down (since y is decreasing) to (0, 0) (where t = 0). We continue to move
to the right (since x is still increasing) but now move upwards (since y is now increasing)
until we reach (1, 2) (where t = 1). Finally, to get a good sense of the shape of the curve, we
√
eliminate the parameter. Solving x = t3 for t, we get t = 3 x. Substituting this into y = 2t2
√
gives y = 2( 3 x)2 = 2x2/3 . Our experience in Section 5.3 yields the graph of our final answer
below.
y
x
1
−1
1
y
2
2
1
1
t
−1
−1
x = t3 , −1 ≤ t ≤ 1
4
1
y = 2t2 , −1 ≤ t ≤ 1
−1
t
˘
1
x
x = t3 , y = 2t2 , −1 ≤ t ≤ 1
You should review Section 1.6.1 if you’ve forgotten what ‘increasing’, ‘decreasing’ and ‘relative minimum’ mean.
11.10 Parametric Equations
1049
2. For the system x = 2e−t , y = e−2t for t ≥ 0, we proceed as in the previous example and
graph x = 2e−t and y = e−2t over the interval [0, ∞). We find that the range of x in this
case is (0, 2] and the range of y is (0, 1]. Next, we plug in some friendly values of t to get a
sense of the orientation of the curve. Since t lies in the exponent here, ‘friendly’ values of t
involve
natural logarithms. Starting with t = ln(1) = 0 we get5 (2, 1), for t = ln(2) we get
1, 41 and for t = ln(3) we get 23 , 19 . Since t is ranging over the unbounded interval [0, ∞),
we take the time to analyze the end behavior of both xand y. As t → ∞, x = 2e−t → 0+
and y = e−2t → 0+ as well. This means the graph of x = 2e−t , y = e−2t approaches the
point (0, 0). Since both x = 2e−t and y = e−2t are always decreasing for t ≥ 0, we know
that our final graph will start at (2, 1) (where t = 0), and move consistently to the left (since
x is decreasing) and down (since y is decreasing) to approach the origin. To
eliminate the
parameter, one way to proceed is to solve x = 2e−t for t to get t = − ln x2 . Substituting
2
2
2
this for t in y = e−2t gives y = e−2(− ln(x/2)) = e2 ln(x/2) = eln(x/2) = x2 = x4 . Or, we could
2
2
2
recognize that y = e−2t = e−t , and sincex = 2e−t means e−t = x2 , we get y = x2 = x4
this way as well. Either way, the graph of x = 2e−t , y = e−2t for t ≥ 0 is a portion of the
2
parabola y = x4 which starts at the point (2, 1) and heads towards, but never reaches,6 (0, 0).
x
y
y
2
1
1
1
x = 2e−t , t ≥ 0
2
t
1
1
y = e−2t , t ≥ 0
2
t
1
˘
2
x
x = 2e−t , y = e−2t , t ≥ 0
3. For the system {x = sin(t), y = csc(t) for 0 < t < π, we start by graphing x = sin(t) and
y = csc(t) over the interval (0, π). We find that the range of x is (0, 1] while the range of
y is [1, ∞). Plotting a few friendly points, we see that t = π6 gives the point 12 , 2 , t = π2
1
gives (1, 1) and t = 5π
6 returns us to 2 , 2 . Since t = 0 and t = π aren’t included in the
domain for t, (because y = csc(t) is undefined at these t-values), we analyze the behavior of
the system as t approaches 0 and π. We find that as t → 0+ as well as when t → π − , we get
x = sin(t) → 0+ and y = csc(t) → ∞. Piecing all of this information together, we get that for
t near 0, we have points with very small
positive x-values, but very large positive y-values.
As t ranges through the interval 0, π2 , x = sin(t) is increasing and y = csc(t)
is decreasing.
1
This means that we are moving to the right and downwards, through 2 , 2 when t = π6 to
(1, 1) when t = π2 . Once t = π2 , the orientation reverses, and we start to head to the left,
since x = sin(t)
and up, since y = csc(t) is now increasing. We pass back
is now decreasing,
5π
1
through 2 , 2 when t = 6 back to the points with small positive x-coordinates and large
5
The reader is encouraged to review Sections 6.1 and 6.2 as needed.
Note the open circle at the origin. See the solution to part 3 in Example 1.2.1 on page 22 and Theorem 4.1 in
Section 4.1 for a review of this concept.
6
1050
Applications of Trigonometry
positive y-coordinates. To better explain this behavior, we eliminate the parameter. Using a
1
reciprocal identity, we write y = csc(t) = sin(t)
. Since x = sin(t), the curve traced out by this
parametrization is a portion of the graph of y = x1 . We now can explain the unusual behavior
as t → 0+ and t → π − – for these values of t, we are hugging the vertical asymptote x = 0 of
the graph of y = x1 . We see that the parametrization given above traces out the portion of
y = x1 for 0 < x ≤ 1 twice as t runs through the interval (0, π).
y
y
3
x
2
1
1
π
π
2
1
t
x = sin(t), 0 < t < π
π
π
2
t
x
1
{x = sin(t), y = csc(t) , 0 < t < π
y = csc(t), 0 < t < π
4. Proceeding as above, we set about graphing {x = 1 + 3 cos(t), y = 2 sin(t) for 0 ≤ t ≤ 3π
2 by
.
We
see
that
x
ranges
first graphing x = 1 + 3 cos(t) and y = 2 sin(t) on the interval 0, 3π
2
from −2 to 4 and y ranges from −2 to 2. Plugging in t = 0, π2 , π and 3π
2 gives the points (4, 0),
(1, 2), (−2, 0) and (1, −2), respectively. As t ranges from 0 to π2 , x = 1 + 3 cos(t) is decreasing,
while y = 2 sin(t) is increasing. This means that we start tracing out our answer at (4, 0) and
continue moving to the left and upwards towards (1, 2). For π2 ≤ t ≤ π, x is decreasing, as is y,
so
the
motion is still right to left, but now is downwards from (1, 2) to (−2, 0). On the interval
3π
π, 2 , x begins to increase, while y continues to decrease. Hence, the motion becomes left
to right but continues downwards, connecting (−2, 0) to (1, −2). To eliminate the parameter
here, we note that the trigonometric functions involved, namely cos(t) and sin(t), are related
by the Pythagorean Identity cos2 (t) + sin2 (t) = 1. Hence, we solve x = 1 + 3 cos(t) for cos(t)
y
to get cos(t) = x−1
3 , and we solve y = 2 sin(t) for sin(t) to get sin(t) = 2 . Substituting these
2
2
2
2
expressions into cos2 (t)+sin2 (t) = 1 gives x−1
+ y2 = 1, or (x−1)
+ y4 = 1. From Section
3
9
7.4, we know that the graph of this equation is an ellipse centered at (1, 0) with vertices at
(−2, 0) and (4, 0) with a minor axis of length 4. Our parametric equations here are tracing
out three-quarters of this ellipse, in a counter-clockwise direction.
x
4
3
y
y
2
2
1
1
−1
π
2
π
3π
2
−2
x = 1 + 3 cos(t), 0 ≤ t ≤
t
−1
2
1
π
2
π
3π
2
y = 2 sin(t), 0 ≤ t ≤
−1
1
2
3
4 x
−1
−2
−2
3π
2
t
3π
2
{x = 1 + 3 cos(t), y = 2 sin(t) , 0 ≤ t ≤
3π
2
11.10 Parametric Equations
1051
Now that we have had some good practice sketching the graphs of parametric equations, we turn
to the problem of finding parametric representations of curves. We start with the following.
Parametrizations of Common Curves
• To parametrize y = f (x) as x runs through some interval I, let x = t and y = f (t) and let
t run through I.
• To parametrize x = g(y) as y runs through some interval I, let x = g(t) and y = t and let
t run through I.
• To parametrize a directed line segment with initial point (x0 , y0 ) and terminal point (x1 , y1 ),
let x = x0 + (x1 − x0 )t and y = y0 + (y1 − y0 )t for 0 ≤ t ≤ 1.
2
2
• To parametrize (x−h)
+ (y−k)
= 1 where a, b > 0, let x = h + a cos(t) and y = k + b sin(t)
a2
b2
for 0 ≤ t < 2π. (This will impart a counter-clockwise orientation.)
The reader is encouraged to verify the above formulas by eliminating the parameter and, when
indicated, checking the orientation. We put these formulas to good use in the following example.
Example 11.10.3. Find a parametrization for each of the following curves and check your answers.
1. y = x2 from x = −3 to x = 2
2. y = f −1 (x) where f (x) = x5 + 2x + 1
3. The line segment which starts at (2, −3) and ends at (1, 5)
4. The circle x2 + 2x + y 2 − 4y = 4
5. The left half of the ellipse
x2
4
+
y2
9
=1
Solution.
1. Since y = x2 is written in the form y = f (x), we let x = tand y = f (t) = t2 . Since x = t, the
bounds on t match precisely the bounds on x so we get x = t, y = t2 for −3 ≤ t ≤ 2. The
check is almost trivial; with x = t we have y = t2 = x2 as t = x runs from −3 to 2.
2. We are told to parametrize y = f −1 (x) for f (x) = x5 + 2x + 1 so it is safe to assume that
f is one-to-one. (Otherwise, f −1 would not exist.) To find a formula y = f −1 (x), we follow
the procedure outlined on page 384 – we start with the equation y = f (x), interchange x and
y and solve for y. Doing so gives us the equation x = y 5 + 2y + 1. While we could attempt
to solve this equation for y, we don’t need to. We can parametrize x = f (y) = y 5 + 2y + 1
by setting y = t so that x = t5 + 2t + 1. We know from our work in Section 3.1 that since
f (x) = x5 + 2x + 1 is an odd-degree polynomial, the range of y = f (x) = x5 + 2x + 1 is
(−∞, ∞). Hence, in order to trace out the entire graph of x = f (y) = y 5 + 2y + 1, we need to
let y run through all real numbers. Our final answer to this problem is x = t5 + 2t + 1, y = t
for −∞ < t < ∞. As in the previous problem, our solution is trivial to check.7
7
Provided you followed the inverse function theory, of course.
1052
Applications of Trigonometry
3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the
formulas x = x0 +(x1 −x0 )t and y = y0 +(y1 −y0 )t for 0 ≤ t ≤ 1. While these equations at first
glance are quite a handful,8 they can be summarized as ‘starting point + (displacement)t’.
To find the equation for x, we have that the line segment starts at x = 2 and ends at x = 1.
This means the displacement in the x-direction is (1 − 2) = −1. Hence, the equation for x is
x = 2 + (−1)t = 2 − t. For y, we note that the line segment starts at y = −3 and ends at
y = 5. Hence, the displacement in the y-direction is (5 − (−3)) = 8, so we get y = −3 + 8t.
Our final answer is {x = 2 − t, y = −3 + 8t for 0 ≤ t ≤ 1. To check, we can solve x = 2 − t
for t to get t = 2 − x. Substituting this into y = −3 + 8t gives y = −3 + 8t = −3 + 8(2 − x),
or y = −8x + 13. We know this is the graph of a line, so all we need to check is that it starts
and stops at the correct points. When t = 0, x = 2 − t = 2, and when t = 1, x = 2 − t = 1.
Plugging in x = 2 gives y = −8(2) + 13 = −3, for an initial point of (2, −3). Plugging in
x = 1 gives y = −8(1) + 13 = 5 for an ending point of (1, 5), as required.
4. In order to use the formulas above to parametrize the circle x2 +2x+y 2 −4y = 4, we first need
to put it into the correct form. After completing the squares, we get (x + 1)2 + (y − 2)2 = 9,
2
2
or (x+1)
+ (y−2)
= 1. Once again, the formulas x = h + a cos(t) and y = k + b sin(t) can be
9
9
a challenge to memorize, but they come from the Pythagorean Identity cos2 (t) + sin2 (t) = 1.
2
2
y−2
In the equation (x+1)
+ (y−2)
= 1, we identify cos(t) = x+1
9
9
3 and sin(t) = 3 . Rearranging
these last two equations, we get x = −1 + 3 cos(t) and y = 2 + 3 sin(t). In order to complete
one revolution around the circle, we let t range through the interval [0, 2π). We get as our final
answer {x = −1 + 3 cos(t), y = 2 + 3 sin(t) for 0 ≤ t < 2π. To check our answer, we could
eliminate the parameter by solving x = −1 + 3 cos(t) for cos(t) and y = 2 + 3 sin(t) for sin(t),
invoking a Pythagorean Identity, and then manipulating the resulting equation in x and y
into the original equation x2 + 2x + y 2 − 4y = 4. Instead, we opt for a more direct approach.
We substitute x = −1 + 3 cos(t) and y = 2 + 3 sin(t) into the equation x2 + 2x + y 2 − 4y = 4
and show that the latter is satisfied for all t such that 0 ≤ t < 2π.
x2 + 2x + y 2 − 4y = 4
?
(−1 + 3 cos(t))2 + 2(−1 + 3 cos(t)) + (2 + 3 sin(t))2 − 4(2 + 3 sin(t)) = 4
?
1 − 6 cos(t) + 9 cos2 (t) − 2 + 6 cos(t) + 4 + 12 sin(t) + 9 sin2 (t) − 8 − 12 sin(t) = 4
?
9 cos2 (t) + 9 sin2 (t) − 5 = 4
?
9 cos2 (t) + sin2 (t) − 5 = 4
?
9 (1) − 5 = 4
X
4 = 4
Now that we know the parametric equations give us points on the circle, we can go through
the usual analysis as demonstrated in Example 11.10.2 to show that the entire circle is covered
as t ranges through the interval [0, 2π).
8
Compare and contrast this with Exercise 65 in Section 11.8.
11.10 Parametric Equations
2
1053
2
5. In the equation x4 + y9 = 1, we can either use the formulas above or think back to the
Pythagorean Identity to get x = 2 cos(t) and y = 3 sin(t). The normal range on the parameter
in this case is 0 ≤ t < 2π, but since we are interested in only the left half of the ellipse, we
restrict t to the values which correspond to Quadrant II and Quadrant III angles, namely
3π
π
3π
π
2 ≤ t ≤ 2 . Our final answer is {x = 2 cos(t), y = 3 sin(t) for 2 ≤ t ≤ 2 . Substituting
2
2
2
2
x = 2 cos(t) and y = 3 sin(t) into x4 + y9 = 1 gives 4 cos4 (t) + 9 sin9 (t) = 1, which reduces
to the Pythagorean Identity cos2 (t) + sin2 (t) = 1. This proves that the points generated by
2
2
the parametric equations {x = 2 cos(t), y = 3 sin(t) lie on the ellipse x4 + y9 = 1. Employing
the techniques demonstrated in Example 11.10.2, we find that the restriction π2 ≤ t ≤ 3π
2
generates the left half of the ellipse, as required.
We note that the formulas given on page 1051 offer only one of literally infinitely many ways to
parametrize the common curves listed there. At times, the formulas offered there need to be altered
to suit the situation. Two easy ways to alter parametrizations are given below.
Adjusting Parametric Equations
• Reversing Orientation: Replacing every occurrence of t with −t in a parametric description for a curve (including any inequalities which describe the bounds on t) reverses
the orientation of the curve.
• Shift of Parameter: Replacing every occurrence of t with (t − c) in a parametric description for a curve (including any inequalities which describe the bounds on t) shifts the
start of the parameter t ahead by c units.
We demonstrate these techniques in the following example.
Example 11.10.4. Find a parametrization for the following curves.
1. The curve which starts at (2, 4) and follows the parabola y = x2 to end at (−1, 1). Shift the
parameter so that the path starts at t = 0.
2. The two part path which starts at (0, 0), travels along a line to (3, 4), then travels along a
line to (5, 0).
3. The Unit Circle, oriented clockwise, with t = 0 corresponding to (0, −1).
Solution.
1. We
y = x2 from x = −1 to x = 2 using the formula given on Page 1051 as
can parametrize
x = t, y = t2 for −1 ≤ t ≤ 2. This parametrization, however, starts at (−1, 1) and ends at
(2, 4). Hence, we need to reverse the orientation. To do so, we replace everyoccurrence of t
with −t to get x = −t, y = (−t)2 for −1 ≤ −t ≤ 2. After simplifying, we get x = −t, y = t2
for −2 ≤ t ≤ 1. We would like t to begin at t = 0 instead of t = −2. The problem here is
that the parametrization we have starts 2 units ‘too soon’, so we need to introduce a ‘time
2
delay’ of 2. Replacing every occurrence
of t with (t −2 2) gives x = −(t − 2), y = (t − 2) for
−2 ≤ t − 2 ≤ 1. Simplifying yields x = 2 − t, y = t − 4t + 4 for 0 ≤ t ≤ 3.
1054
Applications of Trigonometry
2. When parameterizing line segments, we think: ‘starting point + (displacement)t’. For the
first part of the path, we get {x = 3t, y = 4t for 0 ≤ t ≤ 1, and for the second part
we get {x = 3 + 2t, y = 4 − 4t for 0 ≤ t ≤ 1. Since the first parametrization leaves off
at t = 1, we shift the parameter in the second part so it starts at t = 1. Our current
description of the second part starts at t = 0, so we introduce a ‘time delay’ of 1 unit
to the second set of parametric equations. Replacing t with (t − 1) in the second set of
parametric equations gives {x = 3 + 2(t − 1), y = 4 − 4(t − 1) for 0 ≤ t − 1 ≤ 1. Simplifying yields {x = 1 + 2t, y = 8 − 4t for 1 ≤ t ≤ 2. Hence, we may parametrize the path as
{x = f (t), y = g(t) for 0 ≤ t ≤ 2 where
f (t) =
3t, for 0 ≤ t ≤ 1
1 + 2t, for 1 ≤ t ≤ 2
and g(t) =
4t, for 0 ≤ t ≤ 1
8 − 4t, for 1 ≤ t ≤ 2
3. We know that {x = cos(t), y = sin(t) for 0 ≤ t < 2π gives a counter-clockwise parametrization
of the Unit Circle with t = 0 corresponding to (1, 0), so the first order of business is to reverse
the orientation. Replacing t with −t gives {x = cos(−t), y = sin(−t) for 0 ≤ −t < 2π,
which simplifies9 to {x = cos(t), y = − sin(t) for −2π < t ≤ 0. This parametrization gives
a clockwise orientation, but t = 0 still corresponds to the point (1, 0); the point (0, −1) is
reached when t = − 3π
2 . Our strategy is to first get the parametrization to ‘start’ at the
point (0, −1) and then shift the parameter accordingly so the ‘start’ coincides with t = 0.
We know that any interval of length 2π will parametrize the entire circle, so we keep the
equations {x = cos(t), y = − sin(t) , but start the parameter t at − 3π
2 , and find the upper
π
bound by adding 2π so − 3π
≤
t
<
.
The
reader
can
verify
that
{x
= cos(t), y = − sin(t)
2
2
3π
π
for − 2 ≤ t < 2 traces out the Unit Circle clockwise starting at the point (0, −1). We now
every occurrence
shift the parameter
by introducing
a ‘time delay’ of 3π
2 units by
replacing
3π
3π
3π
3π
π
of t with t − 2 . We get x = cos t − 2 , y = − sin t − 2 for − 2 ≤ t − 3π
2 < 2 . This
10
simplifies to {x = − sin(t), y = − cos(t) for 0 ≤ t < 2π, as required.
We put our answer to Example 11.10.4 number 3 to good use to derive the equation of a cycloid.
Suppose a circle of radius r rolls along the positive x-axis at a constant velocity v as pictured below.
Let θ be the angle in radians which measures the amount of clockwise rotation experienced by the
radius highlighted in the figure.
y
P (x, y)
r
θ
x
9
10
courtesy of the Even/Odd Identities
courtesy of the Sum/Difference Formulas
11.10 Parametric Equations
1055
Our goal is to find parametric equations for the coordinates of the point P (x, y) in terms of θ.
From our work in Example 11.10.4 number 3, we know that clockwise motion along the Unit
Circle starting at the point (0, −1) can be modeled by the equations {x = − sin(θ), y = − cos(θ)
for 0 ≤ θ < 2π. (We have renamed the parameter ‘θ’ to match the context of this problem.) To
model this motion on a circle of radius r, all we need to do11 is multiply both x and y by the
factor r which yields {x = −r sin(θ), y = −r cos(θ) . We now need to adjust for the fact that the
circle isn’t stationary with center (0, 0), but rather, is rolling along the positive x-axis. Since the
velocity v is constant, we know that at time t, the center of the circle has traveled a distance vt
down the positive x-axis. Furthermore, since the radius of the circle is r and the circle isn’t moving
vertically, we know that the center of the circle is always r units above the x-axis. Putting these two
facts together, we have that at time t, the center of the circle is at the point (vt, r). From Section
10.1.1, we know v = rθt , or vt = rθ. Hence, the center of the circle, in terms of the parameter θ,
is (rθ, r). As a result, we need to modify the equations {x = −r sin(θ), y = −r cos(θ) by shifting
the x-coordinate to the right rθ units (by adding rθ to the expression for x) and the y-coordinate
up r units12 (by adding r to the expression for y). We get {x = −r sin(θ) + rθ, y = −r cos(θ) + r ,
which can be written as {x = r(θ − sin(θ)), y = r(1 − cos(θ)) . Since the motion starts at θ = 0
and proceeds indefinitely, we set θ ≥ 0.
We end the section with a demonstration of the graphing calculator.
Example 11.10.5. Find the parametric equations of a cycloid which results from a circle of radius
3 rolling down the positive x-axis as described above. Graph your answer using a calculator.
Solution. We have r = 3 which gives the equations {x = 3(t − sin(t)), y = 3(1 − cos(t)) for t ≥ 0.
(Here we have returned to the convention of using t as the parameter.) Sketching the cycloid by
hand is a wonderful exercise in Calculus, but for the purposes of this book, we use a graphing utility.
Using a calculator to graph parametric equations is very similar to graphing polar equations on a
calculator.13 Ensuring that the calculator is in ‘Parametric Mode’ and ‘radian mode’ we enter the
equations and advance to the ‘Window’ screen.
As always, the challenge is to determine appropriate bounds on the parameter, t, as well as for
x and y. We know that one full revolution of the circle occurs over the interval 0 ≤ t < 2π, so
` ´2 ` ´2
If we replace x with xr and y with yr in the equation for the Unit Circle x2 + y 2 = 1, we obtain xr + yr = 1
which reduces to x2 + y 2 = r2 . In the language of Section 1.7, we are stretching the graph by a factor of r in both
the x- and y-directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r.
12
Does this seem familiar? See Example 11.1.1 in Section 11.1.
13
See page 957 in Section 11.5.
11
1056
Applications of Trigonometry
it seems reasonable to keep these as our bounds on t. The ‘Tstep’ seems reasonably small – too
large a value here can lead to incorrect graphs.14 We know from our derivation of the equations of
the cycloid that the center of the generating circle has coordinates (rθ, r), or in this case, (3t, 3).
Since t ranges between 0 and 2π, we set x to range between 0 and 6π. The values of y go from the
bottom of the circle to the top, so y ranges between 0 and 6.
Below we graph the cycloid with these settings, and then extend t to range from 0 to 6π which
forces x to range from 0 to 18π yielding three arches of the cycloid. (It is instructive to note
that keeping the y settings between 0 and 6 messes up the geometry of the cycloid. The reader is
invited to use the Zoom Square feature on the graphing calculator to see what window gives a true
geometric perspective of the three arches.)
14
Again, see page 957 in Section 11.5.
11.10 Parametric Equations
11.10.1
1057
Exercises
In Exercises 1 - 20, plot the set of parametric equations by hand. Be sure to indicate the orientation
imparted on the curve by the parametrization.
1.
x = 4t − 3
for 0 ≤ t ≤ 1
y = 6t − 2
3.
x = 2t
for − 1 ≤ t ≤ 2
y = t2
(
x = t2 + 2t + 1
5.
y =t+1
for t ≤ 1
2.
x = 4t − 1
for 0 ≤ t ≤ 1
y = 3 − 4t
4.
x=t−1
for 0 ≤ t ≤ 3
y = 3 + 2t − t2
(
x=
6.
y=
1
9
1
3t
18 − t2
for t ≥ −3
8.
x = t3
for − ∞ < t < ∞
y=t
10.
x = 3 cos(t)
for 0 ≤ t ≤ π
y = 3 sin(t)
12.
π
x = 3 cos(t)
for ≤ t ≤ 2π
y = 2 sin(t) + 1
2
π
x = 2 cos(t)
for 0 < t <
y = sec(t)
2
14.
π
x = 2 tan(t)
for 0 < t <
y = cot(t)
2
π
π
x = sec(t)
for − < t <
y = tan(t)
2
2
16.
π
3π
x = sec(t)
for < t <
y = tan(t)
2
2
17.
π
π
x = tan(t)
for − < t <
y = 2 sec(t)
2
2
18.
π
3π
x = tan(t)
for < t <
y = 2 sec(t)
2
2
19.
x = cos(t)
for 0 ≤ t ≤ π
y=t
20.
π
π
x = sin(t)
for − ≤ t ≤
y=t
2
2
7.
x=t
for − ∞ < t < ∞
y = t3
9.
π
π
x = cos(t)
for − ≤ t ≤
y = sin(t)
2
2
11.
x = −1 + 3 cos(t)
for 0 ≤ t ≤ 2π
y = 4 sin(t)
13.
15.
In Exercises 21 - 24, plot the set of parametric equations with the help of a graphing utility. Be
sure to indicate the orientation imparted on the curve by the parametrization.
21.
x = t3 − 3t
for − 2 ≤ t ≤ 2
y = t2 − 4
23.
x = et + e−t
for − 2 ≤ t ≤ 2
y = et − e−t
22.
x = 4 cos3 (t)
for 0 ≤ t ≤ 2π
y = 4 sin3 (t)
24.
x = cos(3t)
for 0 ≤ t ≤ 2π
y = sin(4t)
1058
Applications of Trigonometry
In Exercises 25 - 39, find a parametric description for the given oriented curve.
25. the directed line segment from (3, −5) to (−2, 2)
26. the directed line segment from (−2, −1) to (3, −4)
27. the curve y = 4 − x2 from (−2, 0) to (2, 0).
28. the curve y = 4 − x2 from (−2, 0) to (2, 0)
(Shift the parameter so t = 0 corresponds to (−2, 0).)
29. the curve x = y 2 − 9 from (−5, −2) to (0, 3).
30. the curve x = y 2 − 9 from (0, 3) to (−5, −2).
(Shift the parameter so t = 0 corresponds to (0, 3).)
31. the circle x2 + y 2 = 25, oriented counter-clockwise
32. the circle (x − 1)2 + y 2 = 4, oriented counter-clockwise
33. the circle x2 + y 2 − 6y = 0, oriented counter-clockwise
34. the circle x2 + y 2 − 6y = 0, oriented clockwise
(Shift the parameter so t begins at 0.)
35. the circle (x − 3)2 + (y + 1)2 = 117, oriented counter-clockwise
36. the ellipse (x − 1)2 + 9y 2 = 9, oriented counter-clockwise
37. the ellipse 9x2 + 4y 2 + 24y = 0, oriented counter-clockwise
38. the ellipse 9x2 + 4y 2 + 24y = 0, oriented clockwise
(Shift the parameter so t = 0 corresponds to (0, 0).)
39. the triangle with vertices (0, 0), (3, 0), (0, 4), oriented counter-clockwise
(Shift the parameter so t = 0 corresponds to (0, 0).)
40. Use parametric equations and a graphing utility to graph the inverse of f (x) = x3 + 3x − 4.
41. Every polar curve r = f (θ) can be translated to a system of parametric equations with
parameter θ by {x = r cos(θ) = f (θ) cos(θ), y = r sin(θ) = f (θ) sin(θ) . Convert r = 6 cos(2θ)
to a system of parametric equations. Check your answer by graphing r = 6 cos(2θ) by hand
using the techniques presented in Section 11.5 and then graphing the parametric equations
you found using a graphing utility.
42. Use your results from Exercises 3 and 4 in Section 11.1 to find the parametric equations which
model a passenger’s position as they ride the London Eye.
11.10 Parametric Equations
1059
Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force
gravity, the path of the projectile is given by15

 x = v0 cos(θ) t
1
 y = − gt2 + v0 sin(θ) t + s0
2
for 0 ≤ t ≤ T
where v0 is the initial speed of the object, θ is the angle from the horizontal at which the projectile
is launched,16 g is the acceleration due to gravity, s0 is the initial height of the projectile above the
ground and T is the time when the object returns to the ground. (See the figure below.)
y
θ
s0
x
(x(T ), 0)
43. Carl’s friend Jason competes in Highland Games Competitions across the country. In one
event, the ‘hammer throw’, he throws a 56 pound weight for distance. If the weight is released
6 feet above the ground at an angle of 42◦ with respect to the horizontal with an initial speed
of 33 feet per second, find the parametric equations for the flight of the hammer. (Here, use
g = 32 ft.
.) When will the hammer hit the ground? How far away will it hit the ground?
s2
Check your answer using a graphing utility.
44. Eliminate the parameter in the equations for projectile motion to show that the path of the
projectile follows the curve
y=−
g sec2 (θ) 2
x + tan(θ)x + s0
2v02
Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is
y=
15
16
v02 sin2 (θ)
+ s0
2g
when x =
v02 sin(2θ)
2g
A nice mix of vectors and Calculus are needed to derive this.
We’ve seen this before. It’s the angle of elevation which was defined on page 753.
1060
Applications of Trigonometry
45. In another event, the ‘sheaf toss’, Jason throws a 20 pound weight for height. If the weight
is released 5 feet above the ground at an angle of 85◦ with respect to the horizontal and the
sheaf reaches a maximum height of 31.5 feet, use your results from part 44 to determine how
fast the sheaf was launched into the air. (Once again, use g = 32 ft.
.)
s2
46. Suppose θ = π2 . (The projectile was launched vertically.) Simplify the general parametric
formula given for y(t) above using g = 9.8 sm2 and compare that to the formula for s(t) given
in Exercise 25 in Section 2.3. What is x(t) in this case?
In Exercises 47 - 52, we explore the hyperbolic cosine function, denoted cosh(t), and the hyperbolic sine function, denoted sinh(t), defined below:
cosh(t) =
et + e−t
2
and sinh(t) =
et − e−t
2
47. Using a graphing utility as needed, verify that the domain of cosh(t) is (−∞, ∞) and the
range of cosh(t) is [1, ∞).
48. Using a graphing utility as needed, verify that the domain and range of sinh(t) are both
(−∞, ∞).
49. Show that {x(t) = cosh(t), y(t) = sinh(t) parametrize the right half of the ‘unit’ hyperbola
x2 − y 2 = 1. (Hence the use of the adjective ‘hyperbolic.’)
50. Compare the definitions of cosh(t) and sinh(t) to the formulas for cos(t) and sin(t) given in
Exercise 83f in Section 11.7.
51. Four other hyperbolic functions are waiting to be defined: the hyperbolic secant sech(t),
the hyperbolic cosecant csch(t), the hyperbolic tangent tanh(t) and the hyperbolic cotangent
coth(t). Define these functions in terms of cosh(t) and sinh(t), then convert them to formulas
involving et and e−t . Consult a suitable reference (a Calculus book, or this entry on the
hyperbolic functions) and spend some time reliving the thrills of trigonometry with these
‘hyperbolic’ functions.
52. If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise 35 in
Section 6.5, Exercise 47 in Section 6.3 and the answer to Exercise 38 in Section 6.4.
11.10 Parametric Equations
11.10.2
1.
1061
Answers
x = 4t − 3
for 0 ≤ t ≤ 1
y = 6t − 2
2.
x = 4t − 1
for 0 ≤ t ≤ 1
y = 3 − 4t
y
y
4
3
3
2
2
1
1
−3 −2 −1
−1
1
−1
−1
x
1
2
3
x
−2
3.
x = 2t
for − 1 ≤ t ≤ 2
y = t2
4.
x=t−1
for 0 ≤ t ≤ 3
y = 3 + 2t − t2
y
4
4
3
3
2
2
1
1
−3 −2 −1
5.
y
1
2
3
4
−1
x
x = t2 + 2t + 1
for t ≤ 1
y =t+1
6.
1
2
x = 91 18 − t2
y = 31 t
y
2
1
1
1
−2
for t ≥ −3
y
2
−1
x
2
3
4
5
x
−3 −2 −1
−1
1
2
x
1062
7.
Applications of Trigonometry
x=t
for − ∞ < t < ∞
y = t3
8.
x = t3
for − ∞ < t < ∞
y=t
y
y
4
1
3
−4 −3 −2 −1
−1
2
1
2
3
4
x
1
−1
−1
1
x
−2
−3
−4
9.
π
π
x = cos(t)
for − ≤ t ≤
y = sin(t)
2
2
10.
x = 3 cos(t)
for 0 ≤ t ≤ π
y = 3 sin(t)
y
y
1
3
2
1
1 x
−1
−3 −2 −1
1
2
3
x
−1
11.
x = −1 + 3 cos(t)
for 0 ≤ t ≤ 2π
y = 4 sin(t)
12.
π
x = 3 cos(t)
for ≤ t ≤ 2π
y = 2 sin(t) + 1
2
y
y
4
3
3
2
2
1
1
−4 −3 −2 −1
−1
−2
−3
−4
−3
1
2
x
−1
−1
1
3
x
11.10 Parametric Equations
13.
1063
π
x = 2 cos(t)
for 0 < t <
y = sec(t)
2
14.
π
x = 2 tan(t)
for 0 < t <
y = cot(t)
2
y
y
4
4
3
3
2
2
1
1
1
15.
2
3
4
1
x
π
π
x = sec(t)
for − < t <
y = tan(t)
2
2
16.
2
3
π
3π
x = sec(t)
for < t <
y = tan(t)
2
2
y
y
4
4
3
3
2
2
1
1
−1
−4 −3 −2 −1
−1
−2
−2
−3
−3
−4
−4
1
2
3
4
x
4
x
x
1064
17.
Applications of Trigonometry
π
π
x = tan(t)
for − < t <
y = 2 sec(t)
2
2
18.
3π
π
x = tan(t)
for < t <
y = 2 sec(t)
2
2
y
y
4
−2
19.
−2
−1
1
3
−1
2
−2
1
−3
−1
1
2
x = cos(t)
for 0 < t < π
y=t
2
−4
x
20.
π
π
x = sin(t)
for − < t <
y=t
2
2
y
y
π
2
π
−1
π
2
1
x
− π2
−1
21.
x
1
x = t3 − 3t
for − 2 ≤ t ≤ 2
y = t2 − 4
22.
x = 4 cos3 (t)
for 0 ≤ t ≤ 2π
y = 4 sin3 (t)
y
y
−2 −1
−1
−2
−3
−4
1
2
x
4
3
2
1
−4 −3 −2 −1
−1
−2
−3
−4
1 2 3 4 x
x
11.10 Parametric Equations
23.
1065
x = et + e−t
for − 2 ≤ t ≤ 2
y = et − e−t
24.
x = cos(3t)
for 0 ≤ t ≤ 2π
y = sin(4t)
y
y
1
7
5
3
1
−1
1
2
3
4
5
6
7
−1
x
1
x
−3
−5
−1
−7
26.
x = 5t − 2
for 0 ≤ t ≤ 1
y = −1 − 3t
x=t
for − 2 ≤ t ≤ 2
y = 4 − t2
28.
x=t−2
for 0 ≤ t ≤ 4
y = 4t − t2
x = t2 − 9
for − 2 ≤ t ≤ 3
y=t
30.
x = t2 − 6t
for 0 ≤ t ≤ 5
y =3−t
x = 5 cos(t)
for 0 ≤ t < 2π
y = 5 sin(t)
32.
x = 1 + 2 cos(t)
for 0 ≤ t < 2π
y = 2 sin(t)
25.
x = 3 − 5t
for 0 ≤ t ≤ 1
y = −5 + 7t
27.
29.
31.
x = 3 cos(t)
x = 3 cos(t)
for 0 ≤ t < 2π
33.
for 0 ≤ t < 2π
34.
y = 3 − 3 sin(t)
y = 3 + 3 sin(t)
√
x = 1 + 3 cos(t)
x=3+ √
117 cos(t)
for 0 ≤ t < 2π
36.
35.
for 0 ≤ t < 2π
y = sin(t)
y = −1 + 117 sin(t)
x = 2 cos(t)
for 0 ≤ t < 2π
37.
y = 3 sin(t) − 3

 x = 2 cos t − π = 2 sin(t)
2
38.
for 0 ≤ t < 2π
 y = −3 − 3 sin t − π = −3 + 3 cos(t)
2
39. {x(t), y(t) where:
3t, 0 ≤ t ≤ 1
6 − 3t, 1 ≤ t ≤ 2
x(t) =

0, 2 ≤ t ≤ 3


0, 0 ≤ t ≤ 1
4t − 4, 1 ≤ t ≤ 2
y(t) =

12 − 4t, 2 ≤ t ≤ 3


1066
Applications of Trigonometry
40. The parametric equations for the inverse are
41. r = 6 cos(2θ) translates to
x = t3 + 3t − 4
for − ∞ < t < ∞
y=t
x = 6 cos(2θ) cos(θ)
for 0 ≤ θ < 2π.
y = 6 cos(2θ) sin(θ)
42. The parametric equations
which describe
the locations of passengers on the London Eye are
π
π
π
x = 67.5 cos 15 t − 2 = 67.5 sin 15 t
for − ∞ < t < ∞
π
π
y = 67.5 sin 15
t − π2 + 67.5 = 67.5 − 67.5 cos 15
t
x = 33 cos(42◦ )t
for
y = −16t2 + 33 sin(42◦ )t + 6
t ≥ 0. To find when the hammer hits the ground, we solve y(t) = 0 and get t ≈ −0.23 or
1.61. Since t ≥ 0, the hammer hits the ground after approximately t = 1.61 seconds after
it was launched into the air. To find how far away the hammer hits the ground, we find
x(1.61) ≈ 39.48 feet from where it was thrown into the air.
43. The parametric equations for the hammer throw are
v 2 sin2 (85◦ )
v02 sin2 (θ)
+ s0 = 0
+ 5 = 31.5 to get v0 = ±41.34. The initial speed
2g
2(32)
of the sheaf was approximately 41.34 feet per second.
45. We solve y =